Answer 1

The equation of the line can be rewritten as \[x = \frac{y - 6}{2}\] . 

Substituting the value of x in the equation of the circle, we get: \[\left( \frac{y - 6}{2} \right)^2 + y^2 - 2y - 9 = 0\]

\[\Rightarrow \left( y - 6 \right)^2 + 4 y^2 - 8y - 36 = 0\]
\[ \Rightarrow y^2 + 36 - 12y + 4 y^2 - 8y - 36 = 0\]
\[ \Rightarrow 5 y^2 - 20y = 0\]
\[ \Rightarrow y^2 - 4y = 0\]
\[ \Rightarrow y\left( y - 4 \right) = 0\]
\[ \Rightarrow y = 0, 4\]

At y = 0, x = −3
At y = 4, x = −1
Therefore, the coordinates of A and B are

\[\left( - 1, 4 \right) \text{and} \left( - 3, 0 \right)\]

∴ Equation of the circle with AB as its diameter:

\[\left( x + 1 \right)\left( x + 3 \right) + \left( y - 4 \right)\left( y - 0 \right) = 0\]

\[\Rightarrow x^2 + 4x + y^2 - 4y + 3 = 0\]